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EDIT: Preferably a LHS = RHS proof, where you work on one side only then yield the other side.

My way is as follows:

Prove: $\frac{\cos(x)-\cos(2x)}{\sin(x)+\sin(2x)} = \frac{1-\cos(x)}{\sin(x)}$

I use the fact that $\cos(2x)=2\cos^2(x)-1, \sin(2x)=2\sin(x)\cos(x)$

(1) LHS = $\frac{\cos(x)-2\cos^2(x)+1}{\sin(x)(1+2\cos(x))}$

(2) Thus it would suffice to simply prove that $\frac{\cos(x)-2\cos^2(x)+1}{1+2\cos(x)}=1-\cos(x)$

(3) Then I just used simple algebra by letting $u = \cos(x)$ then factorising and simplifying.

(4) Since that equals $1-\cos(x)$ then the LHS = $\frac{1-\cos(x)}{\sin(x)} = $ RHS.

Firstly, on the practice exam, we pretty much only had maximum 2-2.5 minutes to prove this, and this took me some trial and error figuring out which double angle formula to use for cos(2x).

This probably took me 5 minutes just experimenting, and on the final exam there is no way I can spend that long.

What is the better way to do this?

EDIT: I also proved it by multiplying the numerator and denominator by $1-\cos(x)$, since I saw it on the RHS. This worked a lot better, but is that a legitimate proof?

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  • $\begingroup$ Instead of substituting $u=\cos(x)$, then factorising and simplifying, you could multiply $(1+2\cos(x))(1-\cos(x))$ to prove last identity $\endgroup$
    – enzotib
    Aug 3 '20 at 11:37
  • $\begingroup$ @enzotib Yeah I considered that but I need to check with my teacher if that is a valid proof with respect to the guidelines. Usually, we try to simplify one side to get the other, we usually don't move terms across sides. But it might be allowed. $\endgroup$
    – Simplex1
    Aug 3 '20 at 11:49
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$$\frac{\cos x-\cos2x}{\sin x+\sin 2x } = \frac{1-\cos x }{\sin x}\iff \sin x\cos x-\sin x\cos2x=\sin x-\sin x\cos x+$$

$$+\sin2x-\sin2x\cos x\iff \color{red}{\sin x\cos 2x}+\sin x+\sin2x-\color{red}{\sin2x\cos x}-2\sin x\cos x=0\iff$$

$$\color{red}{\sin(-x)}+\sin x=0$$

and we're done by the double implications all through (and assuming the first expression is well defined, of course)

Check all the cancellations are correct and check all the trigonometric identities used above.

Another way: We begin with the left side, again: assuming it is well defined

$$\frac{\cos x-\cos2x}{\sin x+\sin2x}\stackrel{\cos2x=2\cos^2x-1\\\sin2x=2\sin x\cos x}=\frac{\cos x-2\cos^2x+1}{\sin x(1+2\cos x)}\stackrel{-2t^2+t+1=-(2t+1)(t-1)}=$$

$$=\require{cancel}-\frac{\cancel{(2\cos x+1)}(\cos x-1)}{\sin x\cancel{(1+2\cos x)}}\stackrel{\cdot\frac{\cos x+1}{\cos x+1}}=-\frac{\overbrace{(\cos^2x-1)}^{=-\sin^2x}}{\sin x(\cos x+1)}=$$

$$=-\frac{(-\sin x)}{(\cos x+1)}=\frac{\sin x}{\cos x+1}$$

Finally, we show that last right side equals the right side of the original equation:

$$\frac{\sin x}{\cos x+1}\cdot\frac{\cos x-1}{\cos x-1}=\frac{\sin x(\cos x-1)}{\underbrace{\cos^2x-1}_{=-\sin^2x}}=-\frac{\cos x-1}{\sin x}=\frac{1-\cos x}{\sin x}$$

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    $\begingroup$ Is this a legitimate proof though? $\endgroup$
    – Simplex1
    Aug 3 '20 at 11:33
  • $\begingroup$ @Simplex1 Why do you ask? Haven't you worked out this kind of proofs? If you have any doubt, begin by the end and go back to the beginning. Observe the disclaimer at the end: we assume the original expression (the one you need to prove) is well defined. Thusm for example, $\;x\neq 2k\pi,\,\sin x+\sin 2x\neq0\;$ and etc. $\endgroup$
    – DonAntonio
    Aug 3 '20 at 13:12
  • $\begingroup$ Ok I see that it is logically sound since the preposition implies a tautology (-a+a=0). I think my school requires us to use an LHS = RHS proof where you manipulate one side only and it will yield the other side. $\endgroup$
    – Simplex1
    Aug 4 '20 at 5:18
  • $\begingroup$ @Simplex1 That'd be a rather weird requirement...but anyway: I added a new proof. $\endgroup$
    – DonAntonio
    Aug 4 '20 at 7:12
  • $\begingroup$ Thanks for that. This one avoids the unnecessary step of omitting the sin(x) then subbing in u = cos(x). $\endgroup$
    – Simplex1
    Aug 4 '20 at 9:09
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Here's a trigonographic demonstration:

enter image description here

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  • $\begingroup$ Not a proof to conceive in 2-2.5 minutes of the exam, but a wonderful one. $\endgroup$
    – enzotib
    Aug 3 '20 at 13:21
  • $\begingroup$ I would also put in evidence the orange triangle in this copy of your image: drive.google.com/file/d/1e_wnuvau4j0Z12s8HNBKWMZjuSJCQBjB/… $\endgroup$
    – enzotib
    Aug 3 '20 at 13:29
  • $\begingroup$ @enzotib: Thanks for the "wonderful". :) ... As to timing ... The notion for how to represent the left- and right-hand sides of OP's equation occurred to me almost-immediately, and the "ah-ha!" of the parallel lines happened as I drew a reasonably-accurate sketch. (The nice diagram took longer, of course.) All that said, I wouldn't expect someone to do this on an exam. (It often takes a while to find a good trigonograph.) Nevertheless, I like to promote diagrammatic thinking whenever I can. $\endgroup$
    – Blue
    Aug 3 '20 at 13:31
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    $\begingroup$ This is brilliant. Having it all in one picture shows how every side and hence every expression is related. $\endgroup$
    – Simplex1
    Aug 4 '20 at 5:24
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In the domain of your equality the following expression is equivalent:

$\sin(x)(\cos(x)-\cos(2x))=(1-\cos(x))(\sin(x)+\sin(2x)) \iff \\ 2\sin(x)\cos(x)= \sin(x)+\sin(2x)+ \sin(x)\cos(2x)-\cos(x)\sin(2x) \iff \\ 2\sin(x)\cos(x)= \sin(x)+\sin(2x)+\sin(x-2x) \iff 2\sin(x)\cos(x)=\sin(2x) $

This guarantees your result because you move from one expression to other by adding or subtracting equivalent values.

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By the duplication formulas, for $\sin(x)\neq 0$ and $\cos(x)\neq -\frac12$

$$\frac{\cos(x)\color{red}{-\cos(2x)}}{\sin(x)\color{blue}{+\sin(2x)}} =\frac{1-\cos(x)}{\sin(x)}$$

$$\iff \frac{\cos(x)\color{red}{+1-2\cos^2(x)}}{\sin(x)\color{blue}{+2\sin(x)\cos(x)}} =\frac{1-\cos(x)}{\sin(x)}$$

$$\iff \frac{1+\cos(x)-2\cos^2(x)}{1+2\cos(x)} =1-\cos(x)$$

$$\iff 1+\cos(x)-2\cos^2(x) =(1-\cos(x))(1+2\cos (x))$$

which is true.

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If you are familiar with complex number:

Let $z$ be a complex number satisfying $|z|=1$. We want to prove that $z-\frac{1}{z^{2}}=k\left(1-\frac{1}{z}\right)$ where $k$ is real. After some algebraic manipulation, we obtain $k=z+1+\frac{1}{z}$ which is always real.

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